7.1 Integration By Parts/11

$\int {\text{arctan(4t)}}dt$

$u={\text{arctan(4t)}}\qquad dv=dt$

$du={\frac {4}{1+(4t)^{2}}}dt\qquad v=t$

$\int {\text{arctan(4t)}}dt={\text{tarctan(4t)}}-4\int {\frac {t}{1+16t^{2}}}dt={\text{tarctan(4t)}}-{\frac {4}{32}}\int {\frac {1}{u}}={\text{tarctan(4t)}}-{\frac {1}{8}}in(u)={\text{tarctan(4t)}}-{\frac {1}{8}}in(1+16t^{2})+C$

\begin{align} u=1+16t^{2} \\[1ex] du=32t dt \\[1ex] \frac{1}{32}du=t dt

\end{align} </math>